Question 93247
Given:
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{{{(x^2-9)/(2x-6)}}}
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Note that the numerator is the difference of two squares. As such it can be factored
using the following rule:
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{{{A^2 - B^2 = (A-B)*(A+B)}}}
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This form is identical to the numerator in the given expression if you let A = x and B = 3.
Substituting these into the rule you get:
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{{{(x^2 - 9) = x^2 - 3^2 = (x-3)*(x+3)}}}
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So in place of {{{x^2 - 9}}} you can substitute {{{(x-3)*(x+3)}}} to convert the given
expression to:
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{{{((x-3)*(x+3))/(2x-6)}}}
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Then notice that the denominator can be factored because 2 is common to both terms in
the denominator. When you factor the 2 you then have:
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{{{((x-3)*(x+3))/(2(x-3))}}}
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Then you can cancel the term in the denominator that is common with the one in the numerator
to get:
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{{{((cross(x-3))*(x+3))/(2*(cross(x-3)))}}}
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and what remains is:
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{{{(x+3)/2}}}
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This is the answer to the problem.
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Hope this helps you to understand a method for reducing the problem to a lower form.